Operators on a vector space Problem 17 of section 3.D in Linear Algebra Done Right: 
Suppose $V$ is finite-dimensional and $E$ is a subspace of $L(V)$ such that $ST\in E$ and $TS\in E$ for all $S\in L(V)$ and all $T\in E$. Prove that $E={0}$ or $E=L(V)$.
I found a solution here, but would like a better explanation if possible. I'm especially lost on the $E_{ij}$ operators.
 A: Same method, slightly reduced (but I don't think you can go much below, in a proof 'by hand'): 
Let $(e_i)$ be a basis and $(\ell_j)$ a dual basis, i.e. $\ell_j(e_i)=\delta_{ij}$ (Kronecker delta). Suppose $T\in E$ is non-trivial and maps $x\neq 0$ to $y=Tx\neq 0$. Define $S_i\in L(V)$ by 
$$ S_i(u) = x \; \ell_i(u) $$
Let $\tilde{S}_i \in L(V)$ be any map that maps $y$ to $e_i$. Then
$$ M_{ii}=\tilde{S}_i T S_i (e_j) = \tilde{S}_i T (x) \delta_{ij} = \tilde{S}_i (y) \delta_{ij}=e_i \delta_{ij} $$
Each $M_{ii}\in E$, whence also their sum, which is the identity: $M (e_j) = \sum_i M_{ii} (e_j) = e_j $.
It follows that  $E=L(V)$.
Remark: In infinite dimension the proof no longer works and the result is false: The set of compact operators $K(H)$ of a Hilbert space $H$ is a closed two sided ideal in $L(H)$ which is a proper subset when $H$ has infinite dimension.
A: We want to show that, if $E\ne\{0\}$, then $E$ contains the identity operator, because then the property about products implies $E=L(V)$.
Suppose $E\ne\{0\}$. Then there are $T\in E$ and $w\in V$ such that $T(w)=v\ne0$.
Complete $v$ to a basis of $V$, $\{v=v_1,v_2,\dots,v_n\}$.
For $i=1,\dots,n$ let $S_i$ be the operator such that $S_i(v_1)=v_i$ and $S_i(v_k)=0$ for $k=2,\dots,n$. Let also $S'_i$ be the operator such that $S'_i(v_i)=w$ and $S'_i(v_k)=0$, for $k\ne i$.
By assumption, $T_{i}=S_iTS'_i\in E$. Moreover
$$
T_i(v_i)=S_iTS'_i(v_i)=S_iT(w)=S_i(v_1)=v_i
$$
and, if $k\ne i$,
$$
T_i(v_k)=S_iTS'_i(v_k)=0
$$
The operator $T_1+T_2+\dots+T_n$ is the identity and it belongs to $E$, which is closed under sums.

The result is false as soon as $V$ is not finite dimensional. Indeed, consider the subspace of $L(V)$ consisting of the operators having finite dimensional range. This satisfies the required properties and is not the same as $L(V)$.
